A New Method for Research on the Center-Focus Problem of Differential Systems

Abstract

We will introduce Mironenko’s method to discuss the Poincaré center-focus problem, and compare the methods of Lyapunov and Mironenko. We apply the Mironenko method to discuss the qualitative behavior of solutions of some planar polynomial differential systems and derive the sufficient conditions for a critical point to be a center.

1. Introduction

As we know [1, 2], for polynomial differential system 𝑑𝑥𝑑𝑡=−𝑦+^𝑛_𝑖+𝑗=2𝑝_𝑖𝑗𝑥^𝑖𝑦^𝑗,𝑑𝑦𝑑𝑡=𝑥+^𝑛_𝑖+𝑗=2𝑞_𝑖𝑗𝑥^𝑖𝑦^𝑗, (1) where 𝑝_𝑖𝑗 and 𝑞_𝑖𝑗 are real constants, there has been a longstanding problem, called the Poincaré center-focus problem; for the system (1) find explicit conditions of 𝑝_𝑖𝑗 and 𝑞_𝑖𝑗 under which (1) has a center at the origin (0,0); that is, all the orbits nearby are closed. The problem is equivalent to an analogue for a corresponding periodic equation 𝑑𝑟=∑𝑑𝜃^{𝑛−2𝑖=0}𝐴_𝑖(𝜃)𝑟^𝑖1+∑^𝑛−1_𝑖=0𝐵_𝑖(𝜃)𝑟^𝑖𝑟²=𝑅(𝜃,𝑟) (2) to have periodic solutions. To see this let us note that the phase curves of (1) near the origin (0,0) in polar coordinates 𝑥=𝑟cos𝜃, 𝑦=𝑟sin𝜃 are determined by (2), where 𝐴_𝑖(𝜃) and 𝐵_𝑖(𝜃) are polynomials in cos𝜃 and sin𝜃.

Since the closed orbits of (1) correspond to 2𝜋-periodic solutions of (2), the planar vector field (1) has a center at (0, 0) if and only if (2) has a center at 𝑟=0; that is, all the solutions nearby 𝑟=0 are periodic: 𝑟(0)=𝑟(2𝜋) [1, 2].

To discuss the center-focus problem, there are Lyapunov’s method and the others; see the works of Z. Zhang and so forth [2–8]. One of the most commonly used methods is Lyapunov’s method. However, sometimes, the use of this method to calculate the focus quantities is very difficult and the amount of calculation is very large.

In this paper, we apply the method of Mironenko [9] (reflecting function method) to study the qualitative behavior of solutions of (2) and give the sufficient conditions for 𝑟=0 to be a center. We give a few examples to illustrate that, sometimes, Mironenko's method is better than Lyapunov's method.

In the present section, we introduce the concept of the reflecting function, which will be used throughout the rest of this paper.

Consider differential system 𝑥^′=𝑋(𝑡,𝑥),𝑡∈𝑅,𝑥∈𝑅^𝑛, (3) which has a continuously differentiable right-hand side and with a general solution 𝜙(𝑡;𝑡₀,𝑥₀). For each such system, the reflecting function is defined (see Mironenko's [9]) as 𝐹(𝑡,𝑥)∶=𝜙(−𝑡,𝑡,𝑥). Therefore, for any solution 𝑥(𝑡) of (3), we have 𝐹(𝑡,𝑥(𝑡))=𝑥(−𝑡), 𝐹(0,𝑥)=𝑥.

If system (3) is 2𝜔-periodic with respect to 𝑡, and 𝐹(𝑡,𝑥) is its reflecting function, then 𝑇(𝑥)∶=𝐹(−𝜔,𝑥)=𝜙(𝜔;−𝜔,𝑥) is the Poincaré mapping of (3) over the period [−𝜔,𝜔]. Thus, the solution 𝑥=𝜙(𝑡;−𝜔,𝑥₀) of (3) defined on [−𝜔,𝜔] is 2𝜔-periodic if and only if 𝑥₀ is a fixed point of 𝑇(𝑥). The stability of this periodic solution is equivalent to the stability of the fixed point 𝑥₀.

A differentiable function 𝐹(𝑡,𝑥) is a reflecting function of system (3) if and only if it is a solution of the Cauchy problem 𝐹^′_𝑡+𝐹^′_𝑥𝑋(𝑡,𝑥)+𝑋(−𝑡,𝐹)=0,𝐹(0,𝑥)=𝑥. (4)

If 𝐹(𝑡,𝑥) is a reflecting function of system (3), then it is also reflecting function of the following systems: 𝑥^′=𝑋(𝑡,𝑥)+𝐹^′_𝑥⁻¹𝐺(𝑡,𝑥)−𝐺(−𝑡,𝐹(𝑡,𝑥)), (5) where 𝐺(𝑡,𝑥) is an arbitrary vector function. Thus, the shift operators [9, 10] on interval [−𝜔,𝜔] for such systems coincide, and the qualitative behaviors of the periodic solutions of these systems are the same. See papers [11–18] in which reflecting functions are also applied to study the qualitative behavior of solutions of some nonlinear and nonautonomous differential systems.

In the following, we always assume that all equations in this paper have a continuously differentiable right-hand side and have a unique solution for their initial value problem.

2. Main Results

Now, let us consider differential equation (2).

Lemma 1 ([see [9]). Suppose that 𝐹(𝜃,𝑟) is the reflecting function of (2) and 𝐹(𝜃+2𝜋,𝑟)=𝐹(𝜃,𝑟). Then all the solutions of (2) defined on [−𝜋,𝜋] are 2𝜋-periodic; that is, 𝑟=0 is a center.

Lemma 2 ([see [9]). If all the solutions of (2) are 2𝜋-periodic, then its reflecting function is 2𝜋-periodic, too.
First, we introduce the Mironenko’s method. Suppose that 𝐹=𝑟+^∞_𝑖=1𝑎_𝑖(𝜃)𝑟^𝑖 (6) is the reflecting function of (2), where 𝑎_𝑖(0)=0(𝑖=1,2,…). Substituting it into 𝜕𝐹+𝜕𝜃𝜕𝐹𝑅()((𝜕𝑟𝜃,𝑟+𝑅−𝜃,𝐹𝜃,𝑟))=0 (7) and equating the coefficients of like powers of 𝑟, we get the following infinite equations 𝑎^′₁(𝜃)+𝐴₀(𝜃)+𝐴₀()𝑎−𝜃=0;_′2(𝜃)+2𝑎₁(𝜃)𝐴₀(𝜃)+𝐴₀(−𝜃)+𝐴₁(𝜃)+𝐴₁(−𝜃)−𝐴₀(𝜃)𝐵₁(𝜃)−𝐴₀(−𝜃)𝐵₁()⋮−𝜃=0; (8) If 𝑎_𝑖(𝜃+2𝜋)=𝑎_𝑖(𝜃)(𝑖=1,2,…), then 𝐹(𝜃+2𝜋,𝑟)=𝐹(𝜃,𝑟) and 𝑟=0 is a center. If this is a 𝑎_𝑘(𝜃+2𝜋)≠𝑎_𝑘(𝜃) and 𝑎_𝑖(𝜃+2𝜋)=𝑎_𝑖(𝜃)(𝑖=1,2,…,𝑘−1), then 𝑎_𝑘(𝜃)=𝜑_𝑘(𝜃)+ℎ_𝑘, where 𝜑_𝑘(𝜃+2𝜋)=𝜑_𝑘(𝜃),ℎ_𝑘≠0 is a constant and when ℎ_𝑘<0, (0,0) is an unstable focus; when ℎ_𝑘>0, (0,0) is a stable focus.

By the above description, it seems that Mironenko's method is similar to Lyapunov's method, but do not forget that reflecting function has some good symmetry properties, which make our calculations greatly reduced. The following example will illustrate this advantage.

Example 3. Consider system 𝑥̇=−2𝑦+𝑥²−𝑥𝑦−2𝑦²−𝑥𝑦²−𝑦³,𝑦̇=2𝑥+3𝑥𝑦−𝑦²+𝑥𝑦²−𝑦³. (9) Taking 𝑥=𝑟cos𝜃,𝑦=𝑟sin𝜃, it is transformed to equation 𝑑𝑟=()𝑑𝜃cos𝜃−sin𝜃1+𝑟sin𝜃1+(1+𝑟sin𝜃)²𝑟². (10)

Now, we use the method of Lyapunov to discuss is the 𝑟=0 a center or focus? Suppose that 𝑟=𝑐+^∞_𝑖=1𝑏_𝑖(𝜃)𝑐^𝑖,𝑏_𝑖(0)=0,𝑖=1,2,… (11) is a solution of (10). Substituting it into (10), we get 𝑏^′₁(𝜃)+𝑏^′₂(𝜃)𝑐+𝑏^′₃(𝜃)𝑐²+⋯+𝑏^′_𝑛(𝜃)𝑐^𝑛−1=1+⋯2(cos𝜃−sin𝜃)𝜓²−𝑐2sin𝜃cos𝜃𝜓³+𝑐²4sin²𝜃(cos𝜃+sin𝜃)𝜓⁴−𝑐³4sin⁴𝜃𝜓⁵−𝑐⁴8sin⁴𝜃(cos𝜃+sin𝜃)𝜓⁶+𝑐⁵8cos𝜃sin⁵𝜃𝜓⁷+⋯, (12) where 𝜓∶=1+𝑏₁(𝜃)𝑐+𝑏₂(𝜃)𝑐²+⋯+𝑏_𝑛(𝜃)𝑐^𝑛+⋯.

Equating the coefficients of like powers of 𝑐 implies 𝑏^′₁(𝜃)=12(cos𝜃−sin𝜃). (13) Solving it we get 𝑏₁(𝜃)=12(),𝑏sin𝜃+cos𝜃−1_′2(𝜃)=12(cos2𝜃−cos𝜃+sin𝜃−sin𝜃cos𝜃). (14) It implies 𝑏₂(𝜃)=121sin𝜃cos𝜃−sin𝜃−cos𝜃+1−2sin²𝜃,𝑏^′₃(𝜃)=12sin³3𝜃+4cos𝜃−cos²𝜃−sin𝜃−2sin²𝜃cos𝜃+sin²,𝑏𝜃+sin𝜃cos𝜃₃(𝜃)=143sin𝜃+cos𝜃−3sin𝜃cos𝜃−2sin³𝜃+32sin²2𝜃+3cos³5𝜃−3,𝑏^′₄(𝜃)3=−sin𝜃cos𝜃+41cos2𝜃+4sin𝜃cos²𝜃+58cos𝜃sin³𝜃+114sin²3𝜃cos𝜃+8sin⁴𝜃+16cos⁴2𝜃−32cos𝜃+3sin𝜃−138sin²𝜃cos²𝜃+112sin𝜃cos³1𝜃−4cos³3𝜃−4sin³𝜃, (15) so 𝑏₄1=−2sin²1𝜃−3cos³5𝜃+32sin⁴𝜃+sin³𝜃−11112sin𝜃−48cos⁴1𝜃++12cos𝜃1348sin𝜃cos2𝜃+3148sin𝜃cos𝜃+13.48 (16)

From above, we see that the calculation of the expression of 𝑏_𝑖(𝜃)(𝑖=5,6,7,…) will become more difficult. Thus, it is not easy to judge is 𝑟=0 a center?

Now, we try to use the method of Mironenko to discuss is 𝑟=0 a center? In (7) taking ()𝑅=cos𝜃−sin𝜃1+𝑟sin𝜃1+(1+𝑟sin𝜃)²𝑟², (17) we get 𝑎^′₁(𝜃)+𝑎^′₂(𝜃)𝑟+⋯+𝑎^′_𝑛(𝜃)𝑟^𝑛−1++⋯1+2𝑎₁(𝜃)𝑟+3𝑎₂(𝜃)𝑟²+⋯+(𝑛+1)𝑎_𝑛(𝜃)𝑟^𝑛×1+⋯2()−𝑟cos𝜃−sin𝜃2+𝑟sin𝜃cos𝜃₂4sin²𝜃()−𝑟cos𝜃+sin𝜃³4sin⁴𝜃−𝑟⁴8sin⁴𝜃()+𝑟cos𝜃−sin𝜃⁵8cos𝜃sin⁴+1𝜃+⋯2(cos𝜃+sin𝜃)𝜑²+𝑟2cos𝜃sin𝜃𝜑³+𝑟²4sin²𝜃(cos𝜃−sin𝜃)𝜑⁴−𝑟³4sin⁴𝜃𝜑⁵−𝑟⁴8sin⁴𝜃(cos𝜃+sin𝜃)𝜑⁶−𝑟⁵8cos𝜃sin⁵𝜃𝜑⁷+⋯=0. (18) Equating the coefficients of like powers of 𝑟 implies 𝑎^′₁(𝜃)=−cos𝜃,𝑎₁(𝜃)𝑎=−sin𝜃;_′2(𝜃)=2sin𝜃cos𝜃,𝑎₂(𝜃)=sin²𝑎𝜃;_′3(𝜃)=−3sin²𝜃cos𝜃,𝑎₃(𝜃)=−sin³𝑎𝜃;_′4(𝜃)=4sin³𝜃cos𝜃,𝑎₄(𝜃)=sin⁴𝜃; (19) and so forth. Now we can conjecture that 𝑎_𝑖(𝜃)=(−sin𝜃)^𝑖,𝑖=1,2,3,…; that is, 𝐹(𝜃,𝑟)=𝑟/(1+𝑟sin𝜃). It is not difficult to check that 𝐹(𝜃,𝑟)=𝑟/(1+𝑟sin𝜃) is a solution of (7), so it is the reflecting function of (10). As 𝐹(𝜃+2𝜋,𝑟)=𝐹(𝜃,𝑟), so the origin point (0,0) of system (9) is a center.

Such example shows us that, sometimes, the method of Mironenko is better than Lyapunov’s method. Just this advantage is not enough. By the equivalence, we know that 𝐹(𝜃,𝑟)=𝑟/(1+𝑟sin𝜃) also is the reflecting function of the following equations: 𝑑𝑟=𝑑𝜃cos𝜃1+(1+𝑟sin𝜃)²𝑟²+(1+𝑟sin𝜃)²𝐺()𝑟𝜃,𝑟−𝐺−𝜃,,1+𝑟sin𝜃 (20) where 𝐺(𝜃,𝑟) is an arbitrary function. If we take 𝐺()=𝑟𝜃,𝑟^2𝑘−1(1+𝑟sin𝜃)1+(1+𝑟sin𝜃)^2𝑘, (21) where 𝑘 is a positive integer, then (20) becomes 𝑑𝑟=𝑟𝑑𝜃²cos𝜃1+(1+𝑟sin𝜃)^2𝑘−1+𝑟^2𝑘−11+𝑟sin𝜃−𝑟²1+(1+𝑟sin𝜃)^2𝑘. (22) Taking 𝑥=𝑟cos𝜃,𝑦=𝑟sin𝜃, then (22) is transformed to 𝑑𝑥=𝑥𝑑𝑡²−2𝑦−2𝑦²−𝑦³2+2𝑦+𝑦^2𝑘−1+𝑥𝑥²+𝑦^2𝑘−11+𝑦−𝑥²−𝑦²,𝑑𝑦𝑑𝑡=𝑥2+3𝑦+𝑦²2+2𝑦+𝑦^2𝑘−1+𝑦𝑥²+𝑦^2𝑘−11+𝑦−𝑥²−𝑦² (23) and its origin point (0,0) is a center too.

Thus, by using the method of Mironenko we not only solve a center-focus problem, but at the same time open a class of differential equations with the same character of point 𝑟=0. Therefore, we can say, sometimes, the method of Mironenko is more effective than Lyapunov’s method.

Theorem 4. If 𝐹(𝜃,𝑟) is a solution of equation 𝑅(𝜃,𝑟)=𝑅(−𝜃,𝐹) such that 𝐹(0,𝑟)=𝑟 and satisfies (7), then all the solutions of (2) determined on [−𝜋,𝜋] are 2𝜋-periodic; that is, 𝑟=0 is a center.

Proof. By the assumptions, we see that 𝐹(𝜃,𝑟) is a reflecting function of (2) and 2𝜋-periodic; thus by Lemma 1, the present conclusion is correct.

Theorem 5. If there is a function (𝜃,𝑟)(𝜕𝑢/𝜕𝑟≠0), 𝑢(𝜃+2𝜋,𝑟)=𝑢(𝜃,𝑟) which satisfies 𝜕𝑢+𝜕𝜃𝜕𝑢𝑅()(),𝜕𝑟𝜃,𝑟=𝑓𝜃,𝑢 (24) where 𝑓(𝜃,𝑢)+𝑓(−𝜃,𝑢)=0. Then all the solutions of (2) determined on [−𝜋,𝜋] are 2𝜋-periodic; that is, 𝑟=0 is a center.

Proof. Let 𝐹(𝜃,𝑟) be the reflecting function of (2). Now, we prove that it satisfies 𝑢(𝜃,𝑟)=𝑢(−𝜃,𝐹).
In fact, by relation (24), for any solution 𝑟(𝜃) of (2), we have 𝑑𝑢(𝜃,𝑟(𝜃))(((𝜃.𝑑𝜃=𝑓𝜃,𝑢𝜃,𝑟)) (25) In view of 𝑓(𝜃,𝑢)+𝑓(−𝜃,𝑢)=0, we have [9, page 65] 𝑢(𝜃,𝑟(𝜃))=𝑢(−𝜃,𝑟(−𝜃)); (26) that is, 𝑢(𝜃,𝑟)=𝑢(−𝜃,𝐹), (27) which implies that 𝐹(𝜃,𝑟) is a 2𝜋-periodic function with respect to 𝜃. By Lemma 1, the proof of the present theorem is finished.

Corollary 6. If there is a differentiable 2𝜋-periodic function 𝑎(𝜃) such that 𝛼₀∶=𝑎^′(𝜃)+^𝑛_𝑗=0𝑏_𝑗(𝜃)(−𝑎(𝜃))^𝑗,𝛼_𝑘∶=^𝑛_𝑗=𝑘𝑏_𝑗(𝜃)𝐶^𝑘_𝑗(−𝑎(𝜃))^𝑘−𝑗,(𝑘=1,2,…,𝑛) (28)(𝐶^𝑘_𝑛=𝑛!/𝑘!(𝑛−𝑘)!) are 2𝜋-periodic and odd functions, then all the solutions defined on [−𝜋,𝜋] of equation 𝑑𝑟=𝑑𝜃^𝑛_𝑗=0𝑏_𝑗(𝜃)𝑟^𝑗∶=𝑌(𝜃,𝑟) (29) are 2𝜋-periodic.
This result is implied by Theorem 5 if we take 𝑢=𝑎(𝜃)+𝑟 and put 𝜕𝑢+𝜕𝜃𝜕𝑢𝑌()=𝜕𝑟𝜃,𝑟^𝑛_𝑗=0𝛼_𝑗𝑢^𝑗. (30)

Example 7. Taking 𝑎(𝜃)=sin𝜃, we get that all the solutions determined on [−𝜋,𝜋] of the Riccati equation 𝑑𝑟𝑑𝜃=sin²𝜃cos𝜃−cos𝜃+sin𝜃cos𝜃+2sin⁴𝜃𝑟+sin³𝜃𝑟² (31) are 2𝜋-periodic.

Theorem 8. For (2), suppose that ^𝑘_𝑖=0𝐴_𝑖(𝜃)𝐵_𝑘−𝑖(−𝜃)+𝐴_𝑖(−𝜃)𝐵_𝑘−𝑖(𝜃)=0,𝑘=1,2,…,2𝑛−3, (32)𝐴_𝑖(𝜃)=0(𝑖>𝑛−2), 𝐵_𝑗(𝜃)=0(𝑗>𝑛−1), 𝐵₀(𝜃)=1. Then 𝑟=0 is a center.

Proof. By the assumptions, it is not difficult to check that 𝐹(𝜃,𝑟)=𝑟 is the reflecting function of (2), so 𝑟=0 is a center.

Theorem 9. For (2), suppose that ∫^𝜋₀(𝐴₀+𝐴₀)𝑑𝜃=0 and ^𝑘_𝑗=0𝐵_𝑗𝐴₀+𝐴₀−𝐴_𝑗^𝑘−𝑗_𝑖=0𝐵_𝑖𝐶^{𝑘−𝑗−𝑖}_{𝑛−1−𝑖}𝛼^{𝑘−𝑗−𝑖}=^𝑘_𝑗=0𝐵_𝑗+𝐵_𝑗−1𝛼^𝑘−𝑗_𝑖=0𝐴_𝑖𝐶^{𝑘−𝑗−𝑖}_{𝑛−2−𝑖}𝛼^{𝑘−𝑗−𝑖},𝐵𝑘=1,2,…,2𝑛−2,₀=1,𝐴_𝑖(),𝐵=0,𝑖>𝑛−2_𝑗=0,𝑗>𝑛−1, (33) where 𝛼=∫^𝜃₀(𝐴₀+𝐴₀)𝑑𝑠, 𝐴_𝑖=𝐴_𝑖(−𝜃), 𝐵_𝑖=𝐵_𝑖(−𝜃)(𝑖=0,1,2,…,2𝑛−2). Then all the solutions of (2) and the following equations 𝑑𝑟()+()𝑑𝜃=𝑅𝜃,𝑟1+𝑟𝛼²𝐺()𝑟𝜃,𝑟−𝐺−𝜃,1+𝑟𝛼 (34) determined on [−𝜋,𝜋] are 2𝜋-periodic; that is, 𝑟=0 is a center, where 𝐺(𝜃,𝑟) is a continuously differentiable 2𝜋-periodic function with respect to 𝜃 and 𝐺(𝜃,0)=𝐺(−𝜃,0).

Proof. By condition (33), we can check that 𝐹=𝑟/(1+𝑟𝛼) is the reflecting function of (2). As ∫^𝜋₀(𝐴₀+𝐴₀)𝑑𝜃=0, so 𝛼 and 𝐹=𝑟/(1+𝑟𝛼) are 2𝜋-periodic functions with respect to 𝜃. Then by Lemma 1, 𝑟=0 is a center.

Theorem 10. Suppose that for (2) 𝑟=0 is a center and functions Δ_𝑘(𝜃,𝑟) are 2𝜋-periodic with respect to 𝜃 and satisfy Δ_𝜃+Δ_𝑟𝑅(𝜃,𝑟)=𝑅_𝑟(𝜃,𝑟)Δ(𝜃,𝑟),Δ(𝜃,0)=0. (35) Then for the following equations 𝑑𝑟()+𝑑𝜃=𝑅𝜃,𝑟^∞_𝑘=0𝛼_𝑘(𝜃)Δ_𝑘(𝜃,𝑟), (36)𝑟=0 is a center, too. Here 𝛼_𝑘(𝜃)(𝑘=0,1,2,…) are arbitrary continuously differentiable 2𝜋-periodic odd functions.

Proof. By the assumptions and [9, 14], we know that the reflecting functions of (2) and (36) coincide and all the solutions of (2) determined on [−𝜋,𝜋] are 2𝜋-periodic. By Lemma 2, we know that the common reflecting function is 2𝜋-periodic; thus all the solutions of (36) determined on [−𝜋,𝜋] are 2𝜋-periodic; that is, 𝑟=0 is a center, too.

Remark 11. Theorem 10 shows us that if the center-focus problem for (2) is already solved, then the same problem for (36) is solved, too.

Conflict of Interests

The author declares that there is no conflict of interests regarding the publication of this paper.

Acknowledgments

The work supported by the NSF of Jiangsu of China under Grant no. BK2012682 and the NSF of China under Grant nos. 11271026, 61374010.